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Determine the inequality whose solution set is represented by the colored region.
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As the line in our figure is broken or dotted, we know that this corresponds to strictly greater than or strictly less than.
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The reason we use a broken line to represent strict inequalities is because the points on the line itself are not included in the solution set.
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Our first step here will be to find the equation of the line which we know will be in the form π¦ equals ππ₯ plus π, where π is the π¦-intercept and π is the slope or gradient.
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If we select any two points on the line, we can calculate the slope by finding the difference between the π¦-coordinates and dividing this by the difference in the π₯-coordinates.
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This is sometimes known as the rise over the run.
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The slope or gradient π is equal to π¦ two minus π¦ one divided by π₯ two minus π₯ one.
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Whilst it is not essential, it is useful to select two points with integer coordinates.
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In this case, we will choose the two coordinates five, six and negative five, negative nine.
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We will let the first of these have coordinates π₯ one, π¦ one and the second π₯ two, π¦ two.
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Substituting in our values, π is equal to negative nine minus six divided by negative five minus five.
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This simplifies to negative 15 over negative 10.
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And dividing the numerator and denominator by negative five, we see that the slope or gradient of the line is three over two or three-halves.
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This means that our line has equation π¦ is equal to three over two π₯ plus π.
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It appears from the graph that our line intersects the π¦-axis at negative 1.5 or negative three over two.
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We can check this by substituting one of our known points into our equation.
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Substituting in the coordinates five, six gives us six is equal to three over two multiplied by five plus π.
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The right-hand side simplifies to 15 over two or 7.5 plus π.
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Subtracting 15 over two from both sides of our equation, we see that π does indeed equal negative three over two or negative 1.5.
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The π¦-intercept is equal to negative three over two.
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We now have our equation written in slopeβintercept form.
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We see from the figure that all the points in the region that is shaded lie above this line, for example, negative two, eight.
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To determine which of our inequality signs is correct, we can substitute these coordinates into our equation.
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We will then be able to work out whether the left- or right-hand side is greater.
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As π₯ is equal to negative two and π¦ is equal to eight, the left-hand side becomes eight and the right-hand side is equal to three over two multiplied by negative two minus three over two.
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The right-hand side simplifies to negative three minus three over two.
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This is equal to negative nine over two or negative 4.5.
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Eight is clearly larger than this.
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Therefore, the left-hand side is greater than the right-hand side.
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We could repeat this for any other point in our shaded region, telling us that π¦ is greater than three over two π₯ minus three over two.
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Whilst the slopeβintercept form is perfectly acceptable, in this question we will rearrange so that our inequality is in general form.
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We begin by multiplying both sides of the inequality by two.
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Two π¦ is therefore greater than three π₯ minus three.
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Next, we can subtract three π₯ from both sides.
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This gives us the inequality two π¦ minus three π₯ is greater than negative three.
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The inequality whose solution set is represented by the colored region is two π¦ minus three π₯ is greater than negative three.